Finding a closed form for ζ(4) 

And that's a good what? AND THAT'S A GOOD WHAT PROFESSOR?!?

You know you're a math nerd when you know this value by heart.

<a href="#" class="seekto" data-time="105">1:45</a> "Little Twiddle" - thanks! Until now, I did not know the technical term for the "~" symbol. Always a good day when you learn something new!

You can also calculate ζ(4) using Fourier series of f(x) = π^2 - x^2 on [-π, π], and then using Parseval's theorem which relates sum of squares of coefficients with the square of the original function. Parseval might be a cool idea for a future video.

I'll never know if that was a good place to stop or not. The agony.

The good place to stop was obviously before "a good place to stop".

I have used Michael's approach to find a recursive formula for zeta(2*m), where m is any positive integer. Thank you Michael!

Awesome edits! The resulting pace of the video feels great

I love the animation, it's definitely a nice fresh take! Great vid as always:)

and that's a good

I love the new animations throughout the video.

Thank you, professor! That was fun.

@Michael: Could you please do a video on the connection between the values of zeta(2n) and the Bernoulli numbers?

no idea why, but blackpenredpen's picture popping up made me chuckle. great video as always!

extraordinaire et assez bien présenté. Merci pour ces belles perles.

Wonderful. Now could you please do zeta (3)?

It's also possible to obtain ζ(4) Euler's way by "expanding" the 5th order coefficient of the product sin(x)=x*(1-x^2/pi^2)*(1-x^2/(4*pi^2))*... wich is also x^5/120 (taylor serie)

<a href="#" class="seekto" data-time="771">12:51</a> And tha's a good...-----

I would love to see a video on the connection between the Riemann zeta function and the Euler-Mascheroni constant gamma. I've been fascinated with the Riemann zeta function and the Riemann hypothesis for years, and a certain set of videos by 3Blue1Brown over the last decade has only deepened that obsession, but I don't have the depth of formal math education I would like to have.

I belive in Introduction to the Analysis of the Infinities, Euler works out at least up to ζ(12). He uses a different method relating coefficients of polynomials (and by limits, taylor series) with symmetric functions of the roots.

By using repeated integration by parts, this can be generalized to the Fourier series of x^(2n) on [-pi, pi] and it yields a recursive formula for zeta(2n) in terms of all previous zeta(2k) values. If we define z_n to be the coefficient of pi^(2n) in zeta(2n), we get a recursive formula for a sequence of rational numbers related to the Bernoulli numbers.

Hi Dr.! This is so cool!

👍Solutions to S L Loney, Analytical trigonometry compares coefficients in Taylor Expansion, with Product Expansion. Ahlfors (not sure of the author) uses Formal Power Series. Would be nice to have algorithm comparisons on execution difficulties for solving a given problem☺

this method can be used to find higher even numbers of the zeta function.

I paused at <a href="#" class="seekto" data-time="260">4:20</a> and did the calculation myself. The difference was I took x=0 instead of x=pi at the evaluation step. It still works and I got the same answer in the end! However, there was some extra computation... Choosing x=0 turns the cos(nx) into 1 but it doesn't eliminate the (-1)^n . It's easy to fix by splitting the sum into pairs of adjacent odd and even values of n. This led me to a general formula for the alternating version, Sum(n = 1 to inf) of (-1)^n / n^(2k) splitting and re-indexing means it's equal to: Sum(n = 1 to inf) of ( -1/(2n-1)^(2k) + 1/(2n)^(2k) ) In other words, negative of the odd powers, plus the even powers. Set S = zeta(2k) E = just the even powers Factor out the 1/2^(2k) and you just get E = S / 2^(2k) Then the odd powers are S - E So the alternating version = -(S-E) + E = 2E - S = S ( 2 / 2^(2k) - 1) = S * (2^(1-2k) - 1) In other words AlternatingZeta(2k) = Zeta(2k) * (-1 + 1/2^(2k-1)) For example, The Basel problem k=1 is zeta(2) is pi^2 over 6 The alternating version (-1)^n / n^2 Sums to: pi^2 / 6 times (-1 + 1/2^1) or -pi^2 over 12 The alternating version of zeta(4) is -7/8 times pi^4 over 90 Because -1 + 1/8

Another way: we can equate the coefficient of x^4 in the traditional method of finding zeta(2) using the idea: roots of sin x/x. So now, zeta(4)=(zeta(2))^2-2*. So, zeta(4)=pi^4(1/36-1/60)= pi^4/90.

And that's a good

Nice one!

Love it!!

Thanks.

Do the general case with Eisenstein series!

Thank you! ^.^

Absolutely amazing video

The easiest way I found to find zeta(2n) without some sort of formula are the polygamma functions and their reflection formulae. It does require you to take many derivatives though.

That was pretty neat! Can ζ(n) be determined this way by recursion for all other even n?

It would be interesting to focus on the quick resolution of that bn integral, just by assessing the parity of the functions. Would you put a focus on this topic in one of your next videos and why it is so? Thanks!

I did a talk the day before this came out for a closed form formula of zeta(2n)

It was, in fact, not a good place to stop.

The fourier transfor was correct over the -pi to +pi - how did you leap to minus infinity to +infinity required for the zeta ?

What is a good book on doing Fourier series of functions like these?

It seems that you may recover a recursive formula for zeta(2n)?

Great! Now, do with zeta(3)!

Much more interesting would be zeta(3): 😉 Thank you, Michael for your Video.

That was a nice place to stop. Nice video, thank you.

What would be interesting is see where this method fails for f(x)=x^3. A video on that would be very interesting.

I watched the video, but I kept going because I wasn't sure where a good place to stop was.

Outro -> "and" for n -> infinity.

great video it looks like we can do this for all ζ(n) especially for n even but I wonder why it is not going to work for n odd

Values of ζ(2n) are known and there are Bernoulli numbers and π^{2n}

I remember this from Fourier analysis. Why does this fail for odd values again?

Is Zeta(p) proportional to pi^p for any integer p>0 ?

We never knew if this was a good place to st...

What is the reasoning for setting x to pi? RJ

any ideas how to find zeta(3) ?

I wonder if this method is also working for all zeta(n) where n >= 2. like if we can find the Fourier transform of some polynomial of degree m in [-\pi,\pi], in which the sum of the reciprocals of the n-th power appears, then one can calculate zeta(n) using the same procedure shown in the video.

Nice

@BiBenBap did exactly this problem in a video yesterday. Excellent coincidence.

TanQ!

Very good Michael. Why stop there! ζ(8) = ∑ 1/n^8 = π^8/9450 which gives π ~ 3.14153 adding first two terms only which is correct to 4dp Ok I cheated I looked it up in Wolfram.😏

Um, isn't x^4 an even function only when the value of the x-variable itself is even?

Is ok to give the value of x, one of the interval extremety? If we make the Fourier for f(x)=x, we get: x=2*sum [(-1)^(n+1)*(1/n)*sin(nx)] sin(n*pi) is always 0. So if we make x=pi we get pi=2*sum(0)=0 So, pi=0?! In your case is just a luck that for X=pi is the correct result. I think is more correct using Fourier and Parseval Theorem.

Q: The Fourier series converges point-wise on [-\pi,\pi], and not just (-\pi, \pi) because (-\pi)^{4}=\pi^{4}?

<a href="#" class="seekto" data-time="771">12:51</a> 😬🤭

Why is pi showing up everywhere? Where is the circle in this infinite sum?

What's the sin or cos of senpai? ;^}

how did I get here I can't even do long division

At <a href="#" class="seekto" data-time="602">10:02</a> why choose x=pi rather than x=0? Is there something 'wrong' about choosing x=0??

Fire your editor for cutting the ending, mr. Penn 😁

Dear teacher, would you like to solve this problem for me please? Find all number of factor of N that are square integer such that N= 5!8!9!

Brilliant video, but maybe credit should have been given to Euler, who first found this closed form.

incomplete outro <a href="#" class="seekto" data-time="771">12:51</a>

There's a mistake somewhere in calculating a_0, because of the extra factor of 2. Edit: as Star Fox pointed out, my observation was wrong and there are no errors.

<a href="#" class="seekto" data-time="770">12:50</a>

great job！ but it still can't calculate zeta(3) and the series of zeta(4)/[zeta(2)^2] means what ？ 我只是業餘數學愛好者 數學的有趣 就是在於滿足 數學愛好者們對於"calculation"的需求 每個喜歡發表"數學研究成果"的"作者" 都是希望 對數學有興趣的閱覽者 能對該項目"產生興趣" 但語言的隔閡 造成了很多"高等數學內容"失去 很多 原本有興趣的閱覽者

Hi, ok, great! <a href="#" class="seekto" data-time="773">12:53</a> : that's a good.... what?

my 2cents approach, S 1/n4 = (S 1/n2)^2 = (π2/6)^2 = π4/90 obviously 😗 😎

a good what...? a good train? a good neutron star? a good move? .... I'm so confused right now

I don’t understand why you can approximate x⁴ by this formula?